You are given 6 dice (with values 1 to 6) and your goal is to get a total number of pips that is less than 10.
You start by rolling all 6 dice. Say you get $(1,1,2,5,6,6)$. You may (in fact, you must) now choose one number, and put all dice with that number of pips to the side. Say you choose 1, for example. Then $(1,1)$ is put aside. You roll the remaining 4 dice again. Now you might get $(2,4,4,5)$. Again, you choose one number, say 2, and put aside all dice with that number of pips. You now have $(1,1,2)$ on the side and 3 dice left to roll. You repeat this procedure until you can throw no more dice. For instance, if you would now roll the 3 dice and get $(4,4,4)$, we can put all fours to the side and our total score will be the sum of $(1,1,2,4,4,4)$, which is $16$ in this case. The goal is to get a score $<10$.
The question is: what is the best strategy in order to have the highest change of getting below 10? And with this strategy, what is your chance to get a score $<10$?
Some quick simulations show that if your strategy is to always set aside the smallest number that appears on at least one of the rolled dice, then your chance of winning is roughly about 30%. I think there are some cases, though, in which this is not the best strategy. But I haven't seen a nice way to formulate it mathematically.
You can compute your maximum chance of winning, and the corresponding optimal strategy, by using dynamic programming to work backwards from the final throw. Let $\ G(d,t)\ $ denote the game in which you start with $\ d\ $ dice, and win whenever your ultimate total is less than $\ t\ $, and let $\ w(d,t)\ $ be the probability that you win when you play optimally in that game. If you throw your $\ d\ $ dice in that game, the probability that you will get $\ n_1,$$n_2,$$\,\dots,$$n_6\ $ of each of the sides $\ 1,$$2,$$\,\dots,$$6\ $, respectively $\Big($with $\ \displaystyle\sum_{i=1}^6n_i=d\ \Big)$, is $$ p\big(n_1,n_2,\dots,n_6\big)=\frac{{d\choose n_1}{d-n_1\choose n_2}\dots{d-n_1-n_2\dots-n_4\choose n_5}}{6^d}\ . $$ If you now choose to set aside the dice showing side $ i\ $ (where $\ n_i>0\ $), you will then be faced with trying to win the game $\ G(d-n_i,t-in_i)\ $, which you will do with probability $\ w(d-n_i,t-in_i)\ $ if you play optimally from then onwards. If $\ n_i=d\ $ then you win immediately if $\ t-in_i>0\ $, or lose immediately if $\ t-in_i\le0\ $, so we set $\ w(0,t)=1\ $ for $\ t>0 $ and $\ w(0,t)=0\ $ for $\ t\le0 $. Obviously, your optimal first move in $\ G(d,t)\ $ is to choose the value of $\ i\ $ with $\ n_i>0\ $ which maximises $\ w(d-n_i,t-in_i)\ $. Let $\ c^*(\boldsymbol{n})\ $ be the optimal choice of $\ i\ $ for the throw $\ \boldsymbol{n}=\big(n_1,n_2,\dots,n_6\big)\ $. Then $$ w(d,t)=\sum_\boldsymbol{n}p(\boldsymbol{n})w\big(d-n_{c^*(\boldsymbol{n})},c^*(\boldsymbol{n})n_{c^*(\boldsymbol{n})}\big)\ . $$ Once $\ w\big(d',t'\big)\ $ has been tabulated for $\ d'<d\ $ and $\ t'<t\ $, this equation can be used to determine the value of $\ w(d,t)\ $.
The table below gives the values of $\ w(d,t)\ $ for $\ 1\le d\le4\ $ and $\ 2\le t\le11\ $, and approximate values (correct to $7$ decimal places) for $\ d=5,6\ $ and $\ 2\le t\le11\ $. The entry for $\ d=6\ $ and $\ t=10\ $ indicates that your chance of winning the game $\ G(6,10)\ $ is only $17$-$18\%$, which is inconsistent with the OP's statement that your chance of winning it is about $30\%$ if you always choose the smallest number that appears on at least one dice. The table was computed with a script, but I have verified that the entries for $\ d=2\ $ and $\ 3\le t\le7\ $, and $\ d=3\ $ and $\ t=6,7\ $ are correct by hand, so I'm reasonably confident the numbers are all correct. One possible explanation for the discrepancy is that the OP's quick simulations counted a win as occurring when the total was less than or equal to $10$, rather than strictly less than $10$. The table entry for $\ d=6\ $ and $\ t=11\ $ indicates that, with best play, your chances of winning that game are $\ 27$-$28\%\ $. $$ \begin{array}{c|c|c|c|c|c|c|} &2&3&4&5&6&7&8&9&10&11\\ \hline 1&\frac{1}{6}&\frac{1}{3}&\frac{1}{2}&\frac{2}{3}&\frac{5}{6}&1&1&1&1&1\\ \hline 2&0&\frac{2}{27}&\frac{17}{108}&\frac{8}{27}&\frac{23}{54}&\frac{16}{27}&\frac{79}{108}&\frac{23}{27}&\frac{49}{54}&\frac{26}{27}\\ \hline 3&0&0&\frac{163}{3888}&\frac{769}{7776}&\frac{761}{3888}&\frac{2431}{7776}&\frac{875}{1944}&\frac{511}{864}&\frac{5593}{7776}&\frac{6355}{7776}\\ \hline 4&0&0&0&\frac{35387}{2^63^9}&\frac{177793}{2^73^9}&\frac{185101}{2^63^9}&\frac{25565}{2^43^8}&\frac{926741}{2^73^9}&\frac{626611}{2^63^9}&\frac{197237}{2^43^9}\\ \hline 5&0&0&0&0&\tiny0.0210745&\tiny0.0551424&\tiny0.1185422&\tiny0.2025660&\tiny0.3128971&\tiny0.4353564\\ \hline 6&0&0&0&0&0&\tiny0.0170844&\tiny0.0459158&\tiny0.1008971&\tiny0.1759971&\tiny0.2767458\\ \hline \end{array} $$ While there's unlikely to be any simple formula for the optimal strategy for general $\ d\ $ and $\ t\ $, it turns out that for $\ G(6,10)\ $ the best moves are independent of your current position and can be succinctly described by ordering the non-fatal ones from best to worst: $$ \{1,1^2,1^3,1^4,1^5,1^6\}\succ\{2,2^2\}\succ\{3\}\succ\{2^3\}\succ\{4\}\ . $$ Any moves not in the above list will result in a sure loss from any position, and all moves in the above list except the $1$s can become non-viable as the game proceeds. The moves $\ \{2^3\}\ $ and $\ \{4\}\ $, for instance, will only remain viable as long as the dice that have so far been saved are all $1$s.